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Show transcribed image text Note that c = 0 corresponds to the symmetric normal rock-paper-scissors game, in which there is one Nash equilibrium (NE): a mixed-strategy NE where each player plays each of her options with probability 1/3. Suppose you are playing the ummodified, c = 0 game, and that you and your opponent are in equilibrium. Suddenly, the cost of playing rock rises to c = 1 but your opponent does not adjust (i.e., continues to play the (1/3)R + (1/3)P + (1/3)S strategy). Show that the NE of the original game is not an NE of the modified game (recall what is the definition of Nash equilibrium). How should you adjust your strategy now that the game has changed? Find all of your best responses, both pure and mixed, to your opponent's strategy. Suppose each player decides to never play Rock and randomize 50/50 over Paper and Scissors. That is, each plays (1/2)P + (1/2)S. Is this an NE of the c = 1 game? Explain why or why not. Find a symmetric, mixed-strategy NE of the game with c = 1. (Hint: each player plays Rock 1/3 of the time) What is each player's expected utility in the above equilibrium?